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Mathematics Mathematica

Definite integral

Imran Khan

Posted 10 years ago

How to evaluate the following definite integral ??? Integrate[Exp[(Cos[\[Pi] x] - 1)/(2 \[Pi])] Cos[n\[Pi]x], {x, 0, 1}]

POSTED BY: Imran Khan

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Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 10 years ago

This is not a rigorous derivation by any means but it may give a starting point. We generate a list of these for small values of `n`, and work from there. integrals = Table[Integrate[ Exp[(Cos[\[Pi] x] - 1)/(2 \[Pi])] Cos[j\[Pi]x], {x, 0, 1}], {j, 0, 10}] (* Out[51]= {E^(-(1/2)/\[Pi]) BesselI[0, 1/(2 \[Pi])], E^(-(1/2)/\[Pi]) BesselI[1, 1/(2 \[Pi])], E^(-(1/2)/\[Pi]) BesselI[2, 1/(2 \[Pi])], E^(-(1/2)/\[Pi]) BesselI[3, 1/(2 \[Pi])], E^(-(1/2)/\[Pi]) (BesselI[0, 1/(2 \[Pi])] - 4 Hypergeometric0F1Regularized[2, 1/(16 \[Pi]^2)] + 6 Hypergeometric0F1Regularized[3, 1/(16 \[Pi]^2)]), E^(-(1/2)/\[Pi]) (BesselI[1, 1/(2 \[Pi])] + 24 \[Pi] (-BesselI[2, 1/(2 \[Pi])] + 8 \[Pi] BesselI[3, 1/(2 \[Pi])])), E^(-(1/2)/\[Pi]) ((1 + 576 \[Pi]^2 + 30720 \[Pi]^4) BesselI[0, 1/( 2 \[Pi])] - 3 (3 + 512 (\[Pi]^2 + 20 \[Pi]^4)) Hypergeometric0F1Regularized[2, 1/(16 \[Pi]^2)]), E^(-(1/2)/\[Pi]) ((1 + 960 (\[Pi]^2 + 96 \[Pi]^4)) BesselI[1, 1/( 2 \[Pi])] - 48 \[Pi] (1 + 320 (\[Pi]^2 + 48 \[Pi]^4)) BesselI[2, 1/( 2 \[Pi])]), E^(-(1/2)/\[Pi]) ((1 + 1920 (\[Pi]^2 + 240 \[Pi]^4 + 10752 \[Pi]^6)) BesselI[0, 1/( 2 \[Pi])] - 64 \[Pi] (1 + 120 \[Pi]^2 (5 + 192 \[Pi]^2 (3 + 56 \[Pi]^2))) BesselI[1, 1/( 2 \[Pi])]), E^(-(1/2)/\[Pi]) ((1 + 960 \[Pi]^2 (3 + 224 \[Pi]^2 (5 + 384 \[Pi]^2))) BesselI[1, 1/( 2 \[Pi])] - 80 \[Pi] (1 + 192 \[Pi]^2 (5 + 336 \[Pi]^2 (3 + 128 \[Pi]^2))) BesselI[2, 1/( 2 \[Pi])]), E^(-(1/2)/\[Pi]) ((1 + 4800 \[Pi]^2 + 3225600 \[Pi]^4 + 578027520 \[Pi]Out[51]= {E^(-(1/2)/\[Pi]) BesselI[0, 1/(2 \[Pi])], E^(-(1/2)/\[Pi]) BesselI[1, 1/(2 \[Pi])], E^(-(1/2)/\[Pi]) BesselI[2, 1/(2 \[Pi])], E^(-(1/2)/\[Pi]) BesselI[3, 1/(2 \[Pi])], E^(-(1/2)/\[Pi]) (BesselI[0, 1/(2 \[Pi])] - 4 Hypergeometric0F1Regularized[2, 1/(16 \[Pi]^2)] + 6 Hypergeometric0F1Regularized[3, 1/(16 \[Pi]^2)]), E^(-(1/2)/\[Pi]) (BesselI[1, 1/(2 \[Pi])] + 24 \[Pi] (-BesselI[2, 1/(2 \[Pi])] + 8 \[Pi] BesselI[3, 1/(2 \[Pi])])), E^(-(1/2)/\[Pi]) ((1 + 576 \[Pi]^2 + 30720 \[Pi]^4) BesselI[0, 1/( 2 \[Pi])] - 3 (3 + 512 (\[Pi]^2 + 20 \[Pi]^4)) Hypergeometric0F1Regularized[2, 1/(16 \[Pi]^2)]), E^(-(1/2)/\[Pi]) ((1 + 960 (\[Pi]^2 + 96 \[Pi]^4)) BesselI[1, 1/( 2 \[Pi])] - 48 \[Pi] (1 + 320 (\[Pi]^2 + 48 \[Pi]^4)) BesselI[2, 1/( 2 \[Pi])]), E^(-(1/2)/\[Pi]) ((1 + 1920 (\[Pi]^2 + 240 \[Pi]^4 + 10752 \[Pi]^6)) BesselI[0, 1/( 2 \[Pi])] - 64 \[Pi] (1 + 120 \[Pi]^2 (5 + 192 \[Pi]^2 (3 + 56 \[Pi]^2))) BesselI[1, 1/( 2 \[Pi])]), E^(-(1/2)/\[Pi]) ((1 + 960 \[Pi]^2 (3 + 224 \[Pi]^2 (5 + 384 \[Pi]^2))) BesselI[1, 1/( 2 \[Pi])] - 80 \[Pi] (1 + 192 \[Pi]^2 (5 + 336 \[Pi]^2 (3 + 128 \[Pi]^2))) BesselI[2, 1/( 2 \[Pi])]), E^(-(1/2)/\[Pi]) ((1 + 4800 \[Pi]^2 + 3225600 \[Pi]^4 + 578027520 \[Pi]^6 + 23781703680 \[Pi]^8) BesselI[0, 1/( 2 \[Pi])] - 20 \[Pi] (5 + 7680 \[Pi]^2 + 2709504 \[Pi]^4 + 264241152 \[Pi]^6 + 4756340736 \[Pi]^8) BesselI[1, 1/( 2 \[Pi])])} From the above one might guess the general form is `E^(-(1/2)/\[Pi]) BesselI[j, 1/(2 \[Pi])]`. We can test this numerically on the eleven we generated. In[52]:= diffs = integrals - Table[E^(-(1/2)/\[Pi]) BesselI[j, 1/(2 \[Pi])], {j, 0, 10}]; In[53]:= N[diffs, 50] // N During evaluation of In[53]:= N::meprec: Internal precision limit $MaxExtraPrecision = 50.` reached while evaluating {0,0,<<8>>,E^(-(1/2)/\[Pi]) ((1+4800 Power[<<2>>]+3225600 Power[<<2>>]+578027520 Power[<<2>>]+23781703680 Power[<<2>>]) BesselI[0,1/2 Power[<<2>>]]-20 \[Pi] (5+7680 Power[<<2>>]+2709504 Power[<<2>>]+264241152 Power[<<2>>]+4756340736 Power[<<2>>]) BesselI[1,1/2 Power[<<2>>]])-E^(-(1/2)/\[Pi]) BesselI[10,1/(2 \[Pi])]}. >> Out[53]= {0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.}^6 + 23781703680 \[Pi]^8) BesselI[0, 1/( 2 \[Pi])] - 20 \[Pi] (5 + 7680 \[Pi]^2 + 2709504 \[Pi]^4 + 264241152 \[Pi]^6 + 4756340736 \[Pi]^8) BesselI[1, 1/( 2 \[Pi])])} ) In[52]:= diffs = integrals - Table[E^(-(1/2)/\[Pi]) BesselI[j, 1/(2 \[Pi])], {j, 0, 10}]; In[53]:= N[diffs, 50] // N ( During evaluation of In[53]:= N::meprec: Internal precision limit $MaxExtraPrecision = 50.` reached while evaluating {0,0,<<8>>,E^(-(1/2)/\[Pi]) ((1+4800 Power[<<2>>]+3225600 Power[<<2>>]+578027520 Power[<<2>>]+23781703680 Power[<<2>>]) BesselI[0,1/2 Power[<<2>>]]-20 \[Pi] (5+7680 Power[<<2>>]+2709504 Power[<<2>>]+264241152 Power[<<2>>]+4756340736 Power[<<2>>]) BesselI[1,1/2 Power[<<2>>]])-E^(-(1/2)/\[Pi]) BesselI[10,1/(2 \[Pi])]}. >> ) ( Out[53]= {0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.} *)

This is not a rigorous derivation by any means but it may give a starting point. We generate a list of these for small values of n, and work from there.

integrals = 
 Table[Integrate[
   Exp[(Cos[\[Pi] x] - 1)/(2 \[Pi])] Cos[j*\[Pi]*x], {x, 0, 1}], {j, 
   0, 10}]
(* Out[51]= {E^(-(1/2)/\[Pi]) BesselI[0, 1/(2 \[Pi])], 
 E^(-(1/2)/\[Pi]) BesselI[1, 1/(2 \[Pi])], 
 E^(-(1/2)/\[Pi]) BesselI[2, 1/(2 \[Pi])], 
 E^(-(1/2)/\[Pi]) BesselI[3, 1/(2 \[Pi])], 
 E^(-(1/2)/\[Pi]) (BesselI[0, 1/(2 \[Pi])] - 
    4 Hypergeometric0F1Regularized[2, 1/(16 \[Pi]^2)] + 
    6 Hypergeometric0F1Regularized[3, 1/(16 \[Pi]^2)]), 
 E^(-(1/2)/\[Pi]) (BesselI[1, 1/(2 \[Pi])] + 
    24 \[Pi] (-BesselI[2, 1/(2 \[Pi])] + 
       8 \[Pi] BesselI[3, 1/(2 \[Pi])])), 
 E^(-(1/2)/\[Pi]) ((1 + 576 \[Pi]^2 + 30720 \[Pi]^4) BesselI[0, 1/(
      2 \[Pi])] - 
    3 (3 + 512 (\[Pi]^2 + 20 \[Pi]^4)) Hypergeometric0F1Regularized[2,
       1/(16 \[Pi]^2)]), 
 E^(-(1/2)/\[Pi]) ((1 + 960 (\[Pi]^2 + 96 \[Pi]^4)) BesselI[1, 1/(
      2 \[Pi])] - 
    48 \[Pi] (1 + 320 (\[Pi]^2 + 48 \[Pi]^4)) BesselI[2, 1/(
      2 \[Pi])]), 
 E^(-(1/2)/\[Pi]) ((1 + 
       1920 (\[Pi]^2 + 240 \[Pi]^4 + 10752 \[Pi]^6)) BesselI[0, 1/(
      2 \[Pi])] - 
    64 \[Pi] (1 + 
       120 \[Pi]^2 (5 + 192 \[Pi]^2 (3 + 56 \[Pi]^2))) BesselI[1, 1/(
      2 \[Pi])]), 
 E^(-(1/2)/\[Pi]) ((1 + 
       960 \[Pi]^2 (3 + 224 \[Pi]^2 (5 + 384 \[Pi]^2))) BesselI[1, 1/(
      2 \[Pi])] - 
    80 \[Pi] (1 + 
       192 \[Pi]^2 (5 + 336 \[Pi]^2 (3 + 128 \[Pi]^2))) BesselI[2, 1/(
      2 \[Pi])]), 
 E^(-(1/2)/\[Pi]) ((1 + 4800 \[Pi]^2 + 3225600 \[Pi]^4 + 
       578027520 \[Pi]Out[51]= {E^(-(1/2)/\[Pi]) BesselI[0, 1/(2 \[Pi])], 
        E^(-(1/2)/\[Pi]) BesselI[1, 1/(2 \[Pi])], 
        E^(-(1/2)/\[Pi]) BesselI[2, 1/(2 \[Pi])], 
        E^(-(1/2)/\[Pi]) BesselI[3, 1/(2 \[Pi])], 
        E^(-(1/2)/\[Pi]) (BesselI[0, 1/(2 \[Pi])] - 
           4 Hypergeometric0F1Regularized[2, 1/(16 \[Pi]^2)] + 
           6 Hypergeometric0F1Regularized[3, 1/(16 \[Pi]^2)]), 
        E^(-(1/2)/\[Pi]) (BesselI[1, 1/(2 \[Pi])] + 
           24 \[Pi] (-BesselI[2, 1/(2 \[Pi])] + 
              8 \[Pi] BesselI[3, 1/(2 \[Pi])])), 
        E^(-(1/2)/\[Pi]) ((1 + 576 \[Pi]^2 + 30720 \[Pi]^4) BesselI[0, 1/(
             2 \[Pi])] - 
           3 (3 + 512 (\[Pi]^2 + 20 \[Pi]^4)) Hypergeometric0F1Regularized[2,
              1/(16 \[Pi]^2)]), 
        E^(-(1/2)/\[Pi]) ((1 + 960 (\[Pi]^2 + 96 \[Pi]^4)) BesselI[1, 1/(
             2 \[Pi])] - 
           48 \[Pi] (1 + 320 (\[Pi]^2 + 48 \[Pi]^4)) BesselI[2, 1/(
             2 \[Pi])]), 
        E^(-(1/2)/\[Pi]) ((1 + 
              1920 (\[Pi]^2 + 240 \[Pi]^4 + 10752 \[Pi]^6)) BesselI[0, 1/(
             2 \[Pi])] - 
           64 \[Pi] (1 + 
              120 \[Pi]^2 (5 + 192 \[Pi]^2 (3 + 56 \[Pi]^2))) BesselI[1, 1/(
             2 \[Pi])]), 
        E^(-(1/2)/\[Pi]) ((1 + 
              960 \[Pi]^2 (3 + 224 \[Pi]^2 (5 + 384 \[Pi]^2))) BesselI[1, 1/(
             2 \[Pi])] - 
           80 \[Pi] (1 + 
              192 \[Pi]^2 (5 + 336 \[Pi]^2 (3 + 128 \[Pi]^2))) BesselI[2, 1/(
             2 \[Pi])]), 
        E^(-(1/2)/\[Pi]) ((1 + 4800 \[Pi]^2 + 3225600 \[Pi]^4 + 
              578027520 \[Pi]^6 + 23781703680 \[Pi]^8) BesselI[0, 1/(
             2 \[Pi])] - 
           20 \[Pi] (5 + 7680 \[Pi]^2 + 2709504 \[Pi]^4 + 
              264241152 \[Pi]^6 + 4756340736 \[Pi]^8) BesselI[1, 1/(
             2 \[Pi])])}

From the above one might guess the general form is E^(-(1/2)/\[Pi]) BesselI[j, 1/(2 \[Pi])]. We can test this numerically on the eleven we generated.

   In[52]:= diffs = 
     integrals - 
      Table[E^(-(1/2)/\[Pi]) BesselI[j, 1/(2 \[Pi])], {j, 0, 10}];

   In[53]:= N[diffs, 50] // N

   During evaluation of In[53]:= N::meprec: Internal precision limit $MaxExtraPrecision = 50.` reached while evaluating {0,0,<<8>>,E^(-(1/2)/\[Pi]) ((1+4800 Power[<<2>>]+3225600 Power[<<2>>]+578027520 Power[<<2>>]+23781703680 Power[<<2>>]) BesselI[0,1/2 Power[<<2>>]]-20 \[Pi] (5+7680 Power[<<2>>]+2709504 Power[<<2>>]+264241152 Power[<<2>>]+4756340736 Power[<<2>>]) BesselI[1,1/2 Power[<<2>>]])-E^(-(1/2)/\[Pi]) BesselI[10,1/(2 \[Pi])]}. >>

   Out[53]= {0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.}^6 + 23781703680 \[Pi]^8) BesselI[0, 1/(
  2 \[Pi])] - 
20 \[Pi] (5 + 7680 \[Pi]^2 + 2709504 \[Pi]^4 + 
   264241152 \[Pi]^6 + 4756340736 \[Pi]^8) BesselI[1, 1/(
  2 \[Pi])])} *)

In[52]:= diffs = 
  integrals - 
   Table[E^(-(1/2)/\[Pi]) BesselI[j, 1/(2 \[Pi])], {j, 0, 10}];

In[53]:= N[diffs, 50] // N

(* During evaluation of In[53]:= N::meprec: Internal precision limit $MaxExtraPrecision = 50.` reached while evaluating {0,0,<<8>>,E^(-(1/2)/\[Pi]) ((1+4800 Power[<<2>>]+3225600 Power[<<2>>]+578027520 Power[<<2>>]+23781703680 Power[<<2>>]) BesselI[0,1/2 Power[<<2>>]]-20 \[Pi] (5+7680 Power[<<2>>]+2709504 Power[<<2>>]+264241152 Power[<<2>>]+4756340736 Power[<<2>>]) BesselI[1,1/2 Power[<<2>>]])-E^(-(1/2)/\[Pi]) BesselI[10,1/(2 \[Pi])]}. >> *)

(* Out[53]= {0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.} *)

POSTED BY: Daniel Lichtblau

Nasser M. Abbasi

Nasser M. Abbasi, student

Posted 10 years ago

You can't. There is not even an antiderivative to $e^{\cos x}$ never mind the much more complicated one you have! One way these things could be approximated is using series expansion. I'd try numerical integration after plugging some value for `n`. btw, The integral goes to zero quickly as `n` gets large.

POSTED BY: Nasser M. Abbasi

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